public class TestBinaryTree {
    static class TreeNode {
        public char val;
        public TreeNode left;
        public TreeNode right;

        public TreeNode(char val) {
            this.val=val;
        }
    }

    public TreeNode createTree() {
        TreeNode A=new TreeNode('A');
        TreeNode B=new TreeNode('B');
        TreeNode C=new TreeNode('C');
        TreeNode D=new TreeNode('D');
        TreeNode E=new TreeNode('E');
        TreeNode F=new TreeNode('F');
        TreeNode G=new TreeNode('G');
        TreeNode H=new TreeNode('H');

        A.left=B;
        A.right=C;
        B.left=D;
        B.right=E;
        C.left=F;
        C.right=G;
        E.right=H;

        return A;
    }

//    前序遍历
    void preOrder(TreeNode root) {
        if(root==null) return;
        System.out.print(root.val+" ");
        preOrder(root.left);
        preOrder(root.right);
    }

//    中序遍历
    void inOrder(TreeNode root) {
        if(root==null)return;
        inOrder(root.left);
        System.out.print(root.val+" ");
        inOrder(root.right);
    }

//    后序遍历
    void postOrder(TreeNode root) {
        if(root==null)return;
        postOrder(root.left);
        postOrder(root.right);
        System.out.print(root.val+" ");
    }

//    获取树中节点的个数
    // 1. 递归
    public int size(TreeNode root) {
        if(root==null)return 0;
        return size(root.left)+size(root.right)+1;
    }

    // 2. 遍历
    int nodeSize;
    public void size2(TreeNode root) {
        if(root==null)return;
        nodeSize++;
        size2(root.left);
        size2(root.right);
    }

//    求树中叶子节点的个数
    public int leafNodeSize(TreeNode root) {
        if(root==null)return 0;
        if (root.left==null && root.right==null) return 1;
        return leafNodeSize(root.left)+
                leafNodeSize(root.right);
    }

    int leafNodeCount;
    public void leafNodeSize2(TreeNode root) {
        if (root==null) return;
        if (root.left==null && root.right==null) leafNodeCount++;
        leafNodeSize2(root.left);
        leafNodeSize2(root.right);
    }

//    求第k层节点的个数
    public int levelKCount(TreeNode root,int k) {
        if(root==null)return 0;
        if(k==1)return 1;
        /*
        大问题化小问题思想：
        每次递归调用时，k会-1，因为：对于root.left和root.right，它们的子树层级
        比当前层级低1，那么想相应地，目标层级也要-1，才能一直对准，直到递归到目标层级
        为1，这时就开始返回。
        计算该树层级为3的节点个数=A.left(B)的第二层+A.right(C)第二层
        =levelKCount(B, 2)+levelKCount(C, 2)
        =levelKCount(D, 1)+levelKCount(E, 1)+levelKCount(F, 1)+levelKCount(G, 1)
         */
        return levelKCount(root.left,k-1)+
                levelKCount(root.right,k-1);
    }

    public int getHeight(TreeNode root) {
        if(root==null)return 0;
        int leftHeight=getHeight(root.left);
        int rightHeight=getHeight(root.right);
        return leftHeight>rightHeight?leftHeight+1:rightHeight+1;
    }

    public TreeNode find(TreeNode root,char val) {
        if(root==null)return root;
        if(root.val==val)return root;
        TreeNode ret=find(root.left,val);
        if(ret!=null)return ret;

        ret=find(root.right,val);
        if(ret!=null)return ret;

        return null;
    }

}
